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Everything over $\Bbb{C}$. Say we have a smooth curve $C$ of genus $10$ which is a double cover of a smooth plane cubic curve. Therefore $C$ admits a 1-dimensional family of pencils of degree 4 (arising from the involutions on the cubic).

Can we deduce from this that $C$ does not admit any pencil of degree $3$ ? (in other words that $W^1_3( C )$ is empty)

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    $\begingroup$ at least for base point free pencils, how's this? If C admits maps of degrees both 3 and 4 to P^1, then the product map to P^1xP^1 is generically injective, so the genus of C is ≤ 6. The same argument (of Castelnuovo) shows a curve with base point free pencils of relatively prime degrees d,e has genus ≤ (d-1)(e-1). $\endgroup$ – roy smith Jan 9 '16 at 18:00
  • $\begingroup$ technically we need also to rule out a hyperelliptic curve, since then there would be g(1,3)'s with base point. But among the curve of g(1,4)'s there is one whose general divisor does not contain a divisor of the g(1,2) so we get again that the genus is ≤ 3. $\endgroup$ – roy smith Jan 11 '16 at 23:57
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For every integer $p_a>4$, there does not exist a smooth, projective, geometrically connected curve of genus $p_a$ that admits both a degree $2$, finite, flat morphism, $f:C\to E$, to a smooth plane cubic $E$ and a degree $3$, finite, flat morphism, $g:C\to \mathbb{P}^1$, to the projective line. Probably this can be proved directly from the geometric Riemann-Roch theorem, but the argument below just uses the adjunction formula. Note also, a complete intersection in $\mathbb{P}^2\times \mathbb{P}^1$ of hypersurfaces of bidegrees $(3,0)$ and $(1,2)$ is a curve of arithmetic genus $p_a=4$ that does admit a degree $2$ morphism to $E$ and a degree $3$ morphism to $\mathbb{P}^1$. So the inequality $p_a>4$ is necessary.

By way of contradiction, assume that there is such a curve. Consider the product morphism, $$(f,g):C\to E\times\mathbb{P}^1.$$ Denote the image Cartier divisor by $$i:B\to E\times \mathbb{P}^1.$$ This is a closed, integral (possibly singular) curve in $E\times \mathbb{P}^1$. Denote by $$h:C\to B,$$ the unique finite, surjective morphism such that $i\circ h$ equals $(f,g)$.

What is the degree of $h$? For every invertible $\mathcal{O}_E$-module $\mathcal{A}$ of degree $1$, the invertible $\mathcal{O}_C$-module $h^*(i^*\text{pr}_E^*\mathcal{A})$ has degree $2$. Thus, the degree of $h$ divides $2$. Similarly, the degree of $h^*(i^*\text{pr}_{\mathbb{P}^1}^*\mathcal{O}(1))$ has degree $3$. Thus, the degree of $h$ also divides $3$. Therefore the degree of $h$ equals $1$, i.e., $h$ is a birational equivalence. In particular, the arithmetic genus of $B$ is at least as large as the geometric genus $2p_a-2 > 6$ of $C$.

Because $\text{Pic}(E\times \mathbb{P}^1)$ equals $\text{Pic}(E)\times \mathbb{Z}$, there is an isomorphism of invertible sheaves $$\mathcal{O}_{E\times \mathbb{P}^1}(\underline{B}) \cong \text{pr}_E^*\mathcal{L}\otimes \text{pr}_{\mathbb{P}^1}^*\mathcal{O}(d),$$ for an invertible sheaf $\mathcal{L}$ on $E$ and for an integer $d$. By the computations above, $d$ must equal $2$ and $\mathcal{L}$ must have degree $3$. By the adjunction formula, there is an isomorphism of invertible sheaves, $$\omega_B \cong \left( \text{pr}_E^*\mathcal{L} \right)|_B.$$ The degree of this invertible sheaf on $B$ equals $6$, which is strictly less than $2p_a-2$.

Edit. Let $E$ and $F$ be smooth projective curves such that the natural map $\text{Pic}(E)\times \text{Pic}(F) \to \text{Pic}(E\times F)$ is an isomorphism. Let $C$ be a smooth, projective curve. Let $\epsilon:C\to E$ and $\phi:C\to F$ be finite, flat morphisms such that the product morphism $(\epsilon,\phi):C\to E\times F$ is birational to its image, e.g., this holds if $\text{deg}(\epsilon)$ and $\text{deg}(\phi)$ are relatively prime. Then the same argument as above implies the following inequality, $$p_a(C) \leq (\text{deg}(\epsilon)-1)(\text{deg}(\phi)-1) + p_a(E)\text{deg}(\epsilon) + p_a(F)\text{deg}(\phi). $$ In particular, when $F$ is $\mathbb{P}^1$ and $d$ denotes $\text{deg}(\phi)$, so that we are considering $g^1_d$s on $C$, this reduces to the following inequality, $$d \geq \frac{(2p_a(C)-2) - \text{deg}(\epsilon)(2p_a(E)-2)}{2(\text{deg}(\epsilon)-1)} = \frac{\text{deg}(\text{Branch}(\epsilon))}{2(\text{deg}(\epsilon)-1)}.$$ Finally, if also $E$ has genus $1$, this becomes the following, $$d \geq \frac{p_a(C)-1}{\text{deg}(\epsilon)-1}.$$

Second Edit. Thanks to Felipe Voloch who recognized this inequality. It is the Castelnuovo-Severi inequality.

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  • $\begingroup$ Really great answer. Thank you Jason! $\endgroup$ – Heitor Dec 19 '15 at 11:56
  • $\begingroup$ Only, I do not understand the very last formula. Applied to the above example gives $d\geq9$ which is not the case, right? $\endgroup$ – Heitor Dec 19 '15 at 11:59
  • $\begingroup$ @Heitor: "Applied to the above example gives $d\geq 9$ which is not the case, right?" The degree of $\epsilon$ equals $2$, and the degree $d$ of $\phi$ equals $3$. Thus the inequality is $3 \geq (p_a(C)-1)/(2-1)$, i.e., $p_a(C) \leq 4$. So this is the same inequality as in the main part of the argument. $\endgroup$ – Jason Starr Dec 19 '15 at 12:38
  • $\begingroup$ Oh yes, right. My confusion :) Thank you $\endgroup$ – Heitor Dec 19 '15 at 12:41
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    $\begingroup$ AKA Castelnuovo-Severi inequality. $\endgroup$ – Felipe Voloch Dec 19 '15 at 14:44

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